How to Prove Vector Property for R(t) = <f(t), g(t), h(t)>

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The discussion centers on proving the vector property for R(t) = <f(t), g(t), h(t)>, specifically D_t[R(t) X R'(t)] = R(t) X R"(t). Participants clarify that the cross product is denoted by "X" and confirm that the derivative D_t refers to the derivative with respect to t. They emphasize the importance of applying the product rule for vector products, which leads to the realization that the initial equation was miscopied from a textbook. This correction helps participants understand the relationship between the derivatives involved. The conversation concludes with a sense of clarity regarding the proof process.
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I can't seem to figure out how to prove the property for R(t) = <f(t), g(t), h(t)> :

Dt[R(t) X R'(t)] = R(t) X R"(t)

Any suggestions?!
 
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is X cross product?
[R(t) X R'(t)]' = R'(t) X R'(t)+R(t) X R"(t)
for any sensible derivative
thus
[R(t) X R'(t)]' = R(t) X R"(t)
if and only if
R'(t) X R'(t)=0
clearly true for cross product
 
multivariable said:
I can't seem to figure out how to prove the property for R(t) = <f(t), g(t), h(t)> :

Dt[R(t) X R'(t)] = R(t) X R"(t)

Any suggestions?!

Homework Statement





Homework Equations





The Attempt at a Solution


Is the "Dt" derivative with respect to t? i.e.(R x R')' ?

What have you tried? Have you tried actually writing out each side in terms of derivatives of f, g, and h?

Do you know that the "product rule" from Calculus I is still true for vector products? What does that give you?
 
wow.. I had copied the property [R(t) x R'(t)]' = blah blah blah.. incorrectly from my book... It makes perfect sense now, thank you for the help!
 
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